$Pr(X=k)=(\frac{3}{7})(\frac{4}{7})^{k-1}$ for $k=1,2,...$
Enter your answer as an expression in s, where $−\frac{7}{4}<s<\frac{7}{4}$
Also find the mean and variance of $X$
So far I have worked out that $G_X(s)=E[S^X]=\frac{\frac{3}{7}s}{1-\frac{4}{7}s}$ however I am very unsure of this.
In addition the range of values s can take is confusing me as i dont know how to find the mean/variance with this range.
Any help would be massively appreciated!
Your PGF is correct.
The range for $s$ has to do with the fact that the sum of the geometric series $$\sum_{k=0}^\infty z^k = \frac{1}{1-z}$$ is valid only when $|z| < 1$. So if $|s| \ge 7/4$, the PGF does not exist due to the infinite sum diverging.
As for the computation of the expectation and variance, note
$$\frac{dG_X}{ds} = \operatorname{E}[Xs^{X-1}],$$ and setting $s = 1$ gives $$\left[\frac{dG_X}{ds}\right]_{s=1} = \operatorname{E}[X].$$ In other words, the first derivative of the PGF evaluated at $s = 1$ equals the expectation. If we compute the second derivative,
$$\left[\frac{d^2 G_X}{ds^2} \right]_{s=1} = \operatorname{E}[X(X-1)].$$ Then note $$\begin{align}\operatorname{Var}[X] &= \operatorname{E}[X^2] - \operatorname{E}[X]^2 \\[1ex]&= \operatorname{E}[X(X-1) + X] - \operatorname{E}[X]^2 \\[1ex]&= \operatorname{E}[X(X-1)] + \operatorname{E}[X] - \operatorname{E}[X]^2\end{align}$$